1 Inductive Proofs Rosen 6 th ed., §

2 Mathematical Induction A powerful, rigorous technique for proving that a predicate P(n) is true for every natural number n, no matter how large.A powerful, rigorous technique for proving that a predicate P(n) is true for every natural number n, no matter how large. Based on a predicate-logic inference rule:Based on a predicate-logic inference rule: P(0)  n  0 (P(n)  P(n+1))  n  0 P(n)P(0)  n  0 (P(n)  P(n+1))  n  0 P(n)

3 Outline of an Inductive Proof Want to prove  n P(n) …Want to prove  n P(n) … Base case (or basis step): Prove P(0).Base case (or basis step): Prove P(0). Inductive step: Prove  n P(n)  P(n+1).Inductive step: Prove  n P(n)  P(n+1). –e.g. use a direct proof: Let n  N, assume P(n). (inductive hypothesis)Let n  N, assume P(n). (inductive hypothesis) Under this assumption, prove P(n+1).Under this assumption, prove P(n+1). Inductive inference rule then gives  n P(n).Inductive inference rule then gives  n P(n).

4 Induction Example Prove thatProve that

5 Another Induction Example Prove that , n<2 n. Let P(n)=(n<2 n )Prove that , n<2 n. Let P(n)=(n<2 n ) –Base case: P(0)=(0<2 0 )=(0<1)=T. –Inductive step: For prove P(n)  P(n+1). Assuming n<2 n, prove n+1 < 2 n+1.Assuming n<2 n, prove n+1 < 2 n+1. Note n + 1 < 2 n + 1 (by inductive hypothesis) < 2 n + 2 n (because 1<2=2  2   2  2 n-1 = 2 n ) = 2 n+1Note n + 1 < 2 n + 1 (by inductive hypothesis) < 2 n + 2 n (because 1<2=2  2   2  2 n-1 = 2 n ) = 2 n+1 So n + 1 < 2 n+1, and we’re done.So n + 1 < 2 n+1, and we’re done.

6 Validity of Induction Prove: if  n  0 (P(n)  P(n+1)) and P(0), then  k  0 P(k) Prove: if  n  0 (P(n)  P(n+1)) and P(0), then  k  0 P(k) (a) Given any k  0,  n  0 (P(n)  P(n+1)) implies (P(0)  P(1))  (P(1)  P(2))  …  (P(k  1)  P(k)) (P(0)  P(1))  (P(1)  P(2))  …  (P(k  1)  P(k)) Using hypothetical syllogism k-1 times we have (b) Using hypothetical syllogism k-1 times we have P(0)  P(k) P(0)  P(k) (c) P(0) and modus ponens gives P(k). Thus  k  0 P(k). Thus  k  0 P(k).

7 Generalizing Induction Can also be used to prove  n  c P(n) for a given constant c  Z, where maybe c  0, then:Can also be used to prove  n  c P(n) for a given constant c  Z, where maybe c  0, then: –Base case: prove P(c) rather than P(0) –The inductive step is to prove:  n  c (P(n)  P(n+1)).  n  c (P(n)  P(n+1)).

8 Induction Example Prove that the sum of the first n odd positive integers is n 2. That is, prove:Prove that the sum of the first n odd positive integers is n 2. That is, prove: Proof by induction.Proof by induction. –Base case: Let n=1. The sum of the first 1 odd positive integer is 1 which equals 1 2. (Cont…) P(n)P(n)

9 Example cont. Inductive step: Prove  n  1: P(n)  P(n+1).Inductive step: Prove  n  1: P(n)  P(n+1). –Let n  1, assume P(n), and prove P(n+1). By inductive hypothesis P(n)

10 Strong Induction Characterized by another inference rule: P(0)  n  0: (  0  k  n P(k))  P(n+1)  n  0: P(n)Characterized by another inference rule: P(0)  n  0: (  0  k  n P(k))  P(n+1)  n  0: P(n) Difference with previous version is that the inductive step uses the fact that P(k) is true for all smaller, not just for k=n.Difference with previous version is that the inductive step uses the fact that P(k) is true for all smaller, not just for k=n. P is true in all previous cases

11 Example 1 Show that every n>1 can be written as a product p 1 p 2 …p s of some series of s prime numbers. Let P(n)=“n has that property”Show that every n>1 can be written as a product p 1 p 2 …p s of some series of s prime numbers. Let P(n)=“n has that property” Base case: n=2, let s=1, p 1 =2.Base case: n=2, let s=1, p 1 =2. Inductive step: Let n  2. Assume  2  k  n: P(k). Consider n+1. If prime, let s=1, p 1 =n+1. Else n+1=ab, where 1  a  n and 1  b  n. Then a=p 1 p 2 …p t and b=q 1 q 2 …q u. Then n+1= p 1 p 2 …p t q 1 q 2 …q u, a product of s=t+u primes.Inductive step: Let n  2. Assume  2  k  n: P(k). Consider n+1. If prime, let s=1, p 1 =n+1. Else n+1=ab, where 1  a  n and 1  b  n. Then a=p 1 p 2 …p t and b=q 1 q 2 …q u. Then n+1= p 1 p 2 …p t q 1 q 2 …q u, a product of s=t+u primes.

12 Example 2 Prove that every amount of postage of 12 cents or more can be formed using just 4- cent and 5-cent stamps.Prove that every amount of postage of 12 cents or more can be formed using just 4- cent and 5-cent stamps. Base case: 12=3  4), 13=2  4)+1(5), 14=1(4)+2(5), 15=3(5), so  12  n  15, P(n).Base case: 12=3  4), 13=2  4)+1(5), 14=1(4)+2(5), 15=3(5), so  12  n  15, P(n). Inductive step: Let n  15, assume  12  k  n P(k). Note 12  n  3  n, so P(n  3), so add a 4-cent stamp to get postage for n+1.Inductive step: Let n  15, assume  12  k  n P(k). Note 12  n  3  n, so P(n  3), so add a 4-cent stamp to get postage for n+1.

Recursive Definitions Recursion – defining an object (or function, algorithm, etc.) in terms of itselfRecursion – defining an object (or function, algorithm, etc.) in terms of itself Recursion can be used to define sequencesRecursion can be used to define sequences – Previously sequences were defined using a specific formula, e.g., a n = 2 n for n = 0,1,2,... – This sequence can also be defined by giving the first term of the sequence, namely a 0 = 1, and a rule for finding a term of the sequence for the previous one, namely, a n+1 = 2a n for n = 0,1,2,... 13

Recursive Definitions When defining a set recursively, weWhen defining a set recursively, we –(1) specify the initial elements in a basis step and –(2) provide a rule for constructing new elements from those we already have in the recursive step Examples… 14

Recursive Algorithms Sometimes we can reduce the solution to the problem with a particular set of input to the solution of the same problem with smaller input values Sometimes we can reduce the solution to the problem with a particular set of input to the solution of the same problem with smaller input values When such a reduction can be done, the solution to the original problem can be found with a sequence of reductions, until the problem has been reduced to the initial case for which the solution is knownWhen such a reduction can be done, the solution to the original problem can be found with a sequence of reductions, until the problem has been reduced to the initial case for which the solution is known 15

Recursive Algorithms An algorithm is called recursive if it solves a problem by reducing it to an instance of the same problem with smaller input An algorithm is called recursive if it solves a problem by reducing it to an instance of the same problem with smaller input Examples …..Examples ….. 16